﻿ Fundamental Theorem of Calculus Worksheet | Problems & Solutions

# Fundamental Theorem of Calculus Worksheet

Fundamental Theorem of Calculus Worksheet
• Page 1
1.
Use the fundamental theorem of calculus, to evaluate ${\int }_{1}^{2}$($x$4 - 3$x$) $\mathrm{dx}$.
 a. $\frac{9}{2}$ b. $\frac{17}{10}$ c. $\frac{33}{5}$ d. $\frac{107}{10}$ e. $\frac{31}{5}$

#### Solution:

12(x4 - 3x) dx

= 12x4 dx - 3 12x dx
[Use Difference Rule of integration.]

= [x55]12 - 3 [x22]12

= [325-15] - 3 [4 / 2 - 1 / 2]
[Apply the limits of integration.]

= 31 / 5 - 9 / 2

= 62-4510 = 17 / 10

So, 12(x4 - 3x) dx = 17 / 10

2.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{3}$$\frac{1}{4x+3}$ $\mathrm{dx}$.
 a. $\frac{1}{4}$ ln 18 b. $\frac{1}{4}$ ln 5 c. ln 5 d. $\frac{1}{4}$ ln 3 e. $\frac{1}{3}$ ln 5

#### Solution:

0314x+3 dx

= 1 / 4 [ln | 4x + 3 | ]03

= 1 / 4 [ln 15 - ln 3]
[Apply the limits of integration.]

= 1 / 4 ln(15 / 3)

= 1 / 4 ln 5

So, 0314x+3 dx = 1 / 4 ln 5

3.
Use the fundamental theorem of calculus to evaluate, ${\int }_{0}^{1}$($e$3$x$ + $e$ -3$x$) $\mathrm{dx}$.
 a. $\frac{{e}^{3}}{3}$ b. $\frac{{e}^{3}-1}{{e}^{3}}$ c. $\frac{{e}^{3}+1}{3{e}^{3}}$ d. $\frac{{e}^{6}+1}{3{e}^{3}}$ e. $\frac{{e}^{6}-1}{3{e}^{3}}$

#### Solution:

01(e3x + e -3x) dx

= 01e3x dx + 01e -3x dx
[Use Sum Rule of integration.]

= [e3x3]01 + [e-3x-3]01

= e33 - 1 / 3 + e-3-3 + 1 / 3
[Apply the limits of integration.]

= e33-e-33

= e3-e-33 = e6-13e3

So, 01(e3x + e -3x) dx = e6-13e3

4.
Use fundamental theorem of calculus, to evaluate ${\int }_{0}^{\pi /3}$(2 sin $x$ + 3 cos $x$) $\mathrm{dx}$.
 a. $\frac{2-3\sqrt{3}}{2}$ b. $\frac{1+\sqrt{3}}{2}$ c. $\frac{\sqrt{3}}{2}$ d. $\frac{2+3\sqrt{3}}{2}$ e. $\frac{1-\sqrt{3}}{2}$

#### Solution:

0π/3(2 sin x + 3 cos x) dx

= 2 0π/3sin x dx + 3 0π/3cos x dx
[Use Sum Rule of integration.]

= 2 [- cos x ]0π/3 + 3 [sin x ]0π/3

= 2 [- cos π / 3 + cos 0] + 3 [sin π / 3 - sin 0]
[Apply the limits of integration.]

= 2 [- 1 / 2 + 1] + 3 [32]

= 1 + 332 = 2+332

So, 0π/3(2 sin x + 3 cos x) dx = 2+332

5.
Use fundamental theorem of calculus, to evaluate ${\int }_{0}^{1}$$\frac{3}{\sqrt{4-{x}^{2}}}$ $\mathrm{dx}$.
 a. $\frac{1}{2}$ b. $\pi$ c. $\frac{\pi }{2}$ d. $\frac{\pi }{6}$ e. $\frac{3\pi }{2}$

#### Solution:

Let x = 2 sin θ, dx = 2 cosθ dθ

At x = 0, θ = 0 and at x = 1, θ = π6.

0134-x2 dx

= 0π/634-4sin2θ(2 cosθ) dθ

= 0π/66cosθ2cosθ dθ

= 0π/63 dθ

= 3 [ θ ]0π/6
[Apply the limits of integration.]

= 3π6 = π2

So, 0134-x2 dx = π2

6.
Use the fundamental theorem of calculus, to evaluate ${\int }_{0}^{\pi }$$x$ cos2$x$ $\mathrm{dx}$.
 a. ${\pi }^{2}$ b. $\frac{{\pi }^{2}-1}{4}$ c. $\frac{{\pi }^{2}+1}{4}$ d. $\frac{{\pi }^{2}}{4}$ e. $\frac{1}{4}$

#### Solution:

0πx cos2x dx

= 0πx[1+cos2x2] dx

= 0πx2 dx + 0πxcos2x2 dx
[Use Sum Rule of integration.]

= [x24 ]0π + 0πx cos2x2 dx

= π24 + 1 / 2 [[xsin2x2 ]0π - 1 / 2 0πsin 2x dx]
[Use integration by parts.]

= π24 + [xsin2x4 ]0π - 1 / 4 [-cos2x2 ]0π

= π24 + πsin2π4 + 1 / 8 [cos 2π - cos 0]
[Apply the limits of integration.]

= π24 + 0 + 1 / 8 [1 - 1]

= π24

So, 0πx cos2x dx = π24

7.
Use the fundamental theorem of calculus to evaluate ${\int }_{0}^{1}$$\frac{{x}^{2}-1}{{x}^{2}+1}$ $\mathrm{dx}$.
 a. $\frac{2+\pi }{2}$ b. 1 c. $\frac{\pi }{4}$ d. $\frac{2-\pi }{2}$ e. $\frac{1-\pi }{4}$

#### Solution:

01x2-1x2+1 dx

= 01x2+1-2x2+1 dx
[Rearrange the numerator.]

= 01x2+1x2+1 dx - 012x2+1 dx
[Use Difference Rule of integration.]

= 01dx - 2 011x2+12 dx

= [ x ]01 - 2 [tan-1(x1) ]01
[Use 1x2+a2 dx = 1atan-1(xa) + C.]

= 1 - 2Tan -1(1)
[Apply the limits of integration.]

= 1 - π2 = 2-π2

So, 01x2-1x2+1 dx = 2-π2

8.
Use the fundamental theorem of calculus to evaluate ${\int }_{-3}^{0}$ $\frac{x}{{x}^{2}+5}$ $\mathrm{dx}$.
 a. - $\frac{1}{2}$ ln ($\frac{5}{14}$) b. ln ($\frac{5}{14}$) c. $\frac{1}{2}$ ln (14) d. - $\frac{1}{2}$ ln (7) e. $\frac{1}{2}$ ln ($\frac{5}{14}$)

#### Solution:

-30 xx2+5 dx

= 1 / 2 -30 2xx2+5 dx

= 1 / 2 [ln | x2 + 5 | ]-30
[Use f(x)f(x) dx = ln | f(x) | + C.]

= 1 / 2 [ ln 5 - ln 14 ]
[Apply the limits of integration.]

= 1 / 2 ln (5 / 14)

So, -30 xx2+5 dx = 1 / 2 ln (5 / 14)

9.
Use fundamental theorem of calculus to evaluate ${\int }_{1}^{2}$$\frac{1}{\sqrt[3]{1-2x}}$ $\mathrm{dx}$.
 a. $\frac{3}{4}$ b. - $\frac{3}{4}$ ( ${3}^{\frac{2}{3}}$ - 1 ) c. $\frac{3}{4}$ ( ${3}^{\frac{2}{3}}$ + 1 ) d. ( ${3}^{\frac{2}{3}}$ - 1 ) e. $\frac{1}{2}$

#### Solution:

Let 1 - 2x = t, - 2dx = dt, dx = - 1 / 2 dt

At x = 1, t = - 1 and at x = 2, t = - 3

1211-2x3 dx = - 1 / 2 -1 - 31t13 dt

= - 1 / 2 -1 - 3 t-13 dt

= - 1 / 2 [t2323 ]-1- 3
[Apply the limits of integration.]

= - 3 / 4 [(-3)23-(-1)23]

= - 3 / 4 ( 323 - 1)

So, 1211-2x3 dx = - 3 / 4 ( 323 - 1 )

10.
Use fundamental theorem of calculus, to evaluate ${\int }_{0}^{1}$4$x$2(2$x$ - 1)8 $\mathrm{dx}$.
 a. $\frac{1}{11}$ b. $\frac{20}{99}$ c. $\frac{2}{11}$ d. $\frac{1}{9}$ e. $\frac{2}{99}$

#### Solution:

Let 2x - 1 = t, 2 dx = dt, dx = dt2

At x = 0, t = - 1 and at x = 1, t = 1

014x2(2x - 1)8 dx = -114(t+12)2t8dt2

= 1 / 2 -11(t + 1)2t8 dt

= 1 / 2 -11 (t2 + 2t + 1)t8 dt

= 1 / 2 -11 (t10 + 2t9 + t8) dt

= 1 / 2 -11 t10 dt + -11 t9 dt + 1 / 2 -11 t8 dt

= 1 / 2 [t1111 ]-11 + [t1010 ]-11 + 1 / 2 [t99 ]-11

= 1 / 2 [1 / 11 + 1 / 11] + [1 / 10 - 1 / 10] + 12[19+19]
[Apply the limits of integration.]

= 1 / 11 + 1 / 9

= 20 / 99

So, 014x2(2x - 1)8 dx = 20 / 99